Find the value of `1/(81^n)-((10)/(81^n))^(2n)C_1+((10^2)/(81^n))^(2n)C_2-((10^3)/(81^n))^(2n)C_3++(10^(2n))/(81^n)` .

Explanation

To solve the expression

\[ \frac{1}{81^n} - \frac{10}{81^n} \binom{2n}{1} + \frac{10^2}{81^n} \binom{2n}{2} - \frac{10^3}{81^n} \binom{2n}{3} + \ldots + \frac{10^{2n}}{81^n} \]

we can follow these steps:

### Step 1: Factor out the common term
The term \(\frac{1}{81^n}\) is common in all terms of the expression. We can factor it out:

\[ \frac{1}{81^n} \left( 1 - 10 \binom{2n}{1} + 10^2 \binom{2n}{2} - 10^3 \binom{2n}{3} + \ldots + 10^{2n} \right) \]

### Step 2: Recognize the series
The expression inside the parentheses resembles the binomial expansion of \( (1 - x)^{2n} \) where \( x = 10 \):

\[ (1 - 10)^{2n} = \sum_{k=0}^{2n} \binom{2n}{k} (-10)^k \]

This gives us:

\[ 1 - 10 \binom{2n}{1} + 10^2 \binom{2n}{2} - 10^3 \binom{2n}{3} + \ldots + 10^{2n} = (1 - 10)^{2n} \]

### Step 3: Simplify the expression
Now substituting \( (1 - 10)^{2n} \) back into our expression, we have:

\[ \frac{1}{81^n} \left( (1 - 10)^{2n} \right) = \frac{1}{81^n} (-9)^{2n} \]

### Step 4: Calculate \((-9)^{2n}\)
Calculating \((-9)^{2n}\):

\[ (-9)^{2n} = 81^n \]

### Step 5: Substitute back into the expression
Now substituting this back into our expression gives:

\[ \frac{81^n}{81^n} = 1 \]

### Final Answer
Thus, the value of the given expression is:

\[ \boxed{1} \]